$$\frac{1}{u^2}+\frac{1}{v^2}=\frac{1}{w^2}$$
I want to generate all primitive solutions up to $u \le N$. Is there a parametric solution?
By brute force, I got these solutions:
$(15, 20, 12),(20, 15, 12),(65, 156, 60),(136, 255, 120),(156, 65, 60),(175, 600, 168),(255, 136, 120),(580, 609, 420),(600, 175, 168),(609, 580, 420)$.
It seems that $c$ is a multiple of $12$. Another observation is $u^2+v^2=t^4$ where $t$ is the hypotenuse of a triangle.
Wikipedia calls the projective curve $$\{[a:b:c]| a^2b^2=c^2(a^2+b^2)\}$$ the cruciform curve.
Consider the birational map on the projective plane induced by the standard quadratic transformation $$\begin{align}[a:b:c]&=[yz:xz:xy] & [x:y:z]&=[bc:ac:ab]\text{;}\end{align}$$ via this map, the cruciform curve is birational to $$\{[x:y:z] | x^2+y^2=z^2\}\text{,}$$ i.e., the circle. But the circle is birational to the projective line via stereographic projection: $$\begin{align}[x:y:z]&=[t^2-u^2:2tu:t^2+u^2]& [t:u]&=[x+z:y]=[y:z-x]\text{.} \end{align}$$ Consequently, the cruciform curve is also birational to the projective line: $$\begin{align}[a:b:c]&=[2tu(t^2+u^2):(t^2-u^2)(t^2+u^2):2tu(t^2-u^2)]\\ [t:u]&=[b(a+c):ac]=[ac:b(a-c)] \text{.}\end{align}$$